Statement and notation
Let L be [0,2] x [0,2] with the open top-right unit square removed. Four closed discs of radius r must lie in L with disjoint interiors. Write r* for the number in the theorem. It satisfies
r* = 0.377016124861494…
(6 + 4√2)r*² − (12 + 4√2)r* + 5 = 0.
The interval 0.3770161248 < r* < 0.3770161249 follows directly by squaring rational bounds for the radicals. It will also make the few strict diameter comparisons below exact rational checks, not floating-point assumptions.
Construction: the value is attained
Put t = (1 + √2)r* and choose the four centres
A = (1 − r*, 2 − r*), B = (r*, t),
C = (t, r*), D = (2 − r*, 1 − r*).
A and D touch the outside boundary and the two sides of the notch. B and C clear the notch because 1-r* > r*. The middle contact is
|BC| = √2(t − r*) = 2r*.
The quadratic above is exactly |AB| = |CD| = 2r*. For the other pairs, A-C and B-D already differ by 2 − 2r* in one coordinate, while |AD| = √2; all three distances exceed 2r*. Thus all four discs fit, and the optimum is at least r*.
Where four centres are forced to live
Four disjoint discs have total area 4πr² ≤ area(L) = 3, so r ≤ √(3/(4π)) < 1/2. For an upper bound suppose further that r ≥ r*. A centre is in one of three pieces: the upper arm V, the right arm H, or the lower-left central square C. The upper arm is the rectangle
V = [r, 1 − r] × [1, 2 − r],
and H is its reflection in x = y. Their common diameter is √((1 − 2r)² + (1 − r)²), which is below 2r once r > 3 − √7. Since r* is larger than that threshold, V and H hold at most one centre each.
Split C along x = y. In its upper half, y ≥ x and clearance from the notch corner gives √2(1 − x) ≥ r, hence x ≤ a(r) = 1 − r/√2. That half has diameter at most
√((1 − r)² + (a(r) − r)²).
At r* this is below 0.718, while 2r* is above 0.754; the left side decreases with r. The reflected half obeys the same bound. Thus C holds at most two mutually separated centres. Four discs therefore force exactly one centre in V, two in C on opposite sides of the diagonal, and one in H.
The surviving branch
Call the central centre above the diagonal B=(x,y). Reflect the central centre below the diagonal, and call the reflected point E=(s,t). Both have second coordinate at least the first. Each must be at distance at least 2r from a centre in V. For a fixed point below V, the farthest point of V is one of its two top corners.
Class I consists of points with x ≤ 1/2. Their farther corner is (1 − r, 2 − r). Since its horizontal gap is at most 1 − 2r, separation by 2r forces
y ≤ g(r) := 2 − r − √(4r − 1).
Class II has x ≥ 1/2. Notch clearance and y ≥ x give x ≤ a(r). Two class-II points, after reflecting one, are at distance at most 1/√2, below 2r*. A mixed pair is at distance at most
√((1 − r)² + M(r)²),
M(r) = max(g(r) − 1/2, a(r) − r).
Both terms in M decrease with r. At r* this bound is below 0.746, again below 2r* > 0.754. So neither class II nor a mixed pair can be the two central centres. Both B and E are class I.
Every coordinate of B and E now lies in [r,g(r)]. Reflecting E does not change that square, so the two actual central centres are at distance at most its diagonal:
distance ≤ √2(g(r) − r).
Solving the last inequality
Non-overlap requires 2r ≤ √2(g(r) − r). The right side minus 2r is strictly decreasing for r > 1/4. Equality is
√(4r − 1) = 2 − (2 + √2)r,
and squaring gives (6 + 4√2)r² − (12 + 4√2)r + 5 = 0. Its only root in the relevant interval is r*. Therefore r ≤ r*. Together with the construction, the optimum is exactly r*. ∎
What the verifier can record
P08 accepts at most nine decimal places. Since 0.377016124 < r* < 0.377016125, no valid certificate can score the next grid value. Cphysics's standing certificate scores 0.377016124 and is therefore optimal on the site's exact integer verifier.
Cphysics #36 — attaining construction and standing record